(v) Solve the quadratic equation: $7x + 1 = 6x^2$
Solution:
The given equation is:
$7x + 1 = 6x^2$
First, we rearrange the terms to bring the equation into the standard quadratic form $ax^2 + bx + c = 0$:
$0 = 6x^2 - 7x - 1$
Or, more conventionally:
$6x^2 - 7x - 1 = 0$
Comparing this with the standard quadratic equation $ax^2 + bx + c = 0$, we identify the coefficients:
$b = -7$
$c = -1$
Next, we calculate the discriminant, $D = b^2 - 4ac$:
$D = (-7)^2 - 4(6)(-1)$
$D = 49 - (-24)$
$D = 49 + 24$
$D = 73$
Since the discriminant $D > 0$, there are two distinct real roots. We use the quadratic formula to find the roots of $x$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Substituting the values of $a, b, c$ (or $b$ and $D$):
$$x = \frac{-(-7) \pm \sqrt{73}}{2(6)}$$
$$x = \frac{7 \pm \sqrt{73}}{12}$$
Therefore, the two roots of the equation are:
$x_1 = \frac{7 + \sqrt{73}}{12}$
$x_2 = \frac{7 - \sqrt{73}}{12}$